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## Floor Systems & Framing of Floors

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**Floor Systems & Framing of Floors**Residential Architectural Drafting**Design Criteria for Structural Loading**• Load Types • Dead loads • Live loads • Dynamic loads**Dead Loads**• Definition: (DL) loads that make up the actual weight of the structure, such as walls, floors, roofs and any permanently fixed loads such as furnace, air conditioner or other service equipment. Materials that make up the walls, such as, studs, plywood, insulation, sheet rock, nails, glue, etc. are DL. • Building codes specify a minimum of 10#/sq ft for floors and ceilings, DL = 10#/sqft**Live Loads**• Definition: (LL) loads that are fluctuating and changing through the use of the building. These loads include: people, furniture, and exterior weather related items such as, ice, snow, rain, etc. • Building codes specify the amount of live load upon type of use or occupancy. Codes differ, common residential LL = 40#/sq ft**Dynamic Loads**• Definition: loads imposed on the structure by outside natural forces, such as wind and earthquake. • Wind loads • Shear wall design used to resist wind pressure • Uplift forces placed upon the roof • Earthquake loads • Seismic loads causing lateral forces on entire structure**Framing Spacing Practices**• Code based, acceptance varies • Spacing Options • 12” OC • 16” OC • 19.2” OC • 24” OC Common Spacing**12”**15’-0” SPAN 16” 12’-0” SPAN 19.2” 18’-0” SPAN 24” 10’-0” SPAN Load Consideration for one Joist 1’ X 15’ = 15 SQFT X 50 = 750# • Considering 12” OC joist spacing • Considering 16” OC joist spacing • Considering 19.2” OC joist spacing • Considering 24” OC joist spacing 1.33’ X 12’ = 15.96 SQFT X 50 = 798# 1.6’ X 18’ = 28.8 SQFT X 50 = 1440# 2’ X 10’ = 20 SQFT X 50 = 1000#**Sizing Joist Using Span Tables**• Loading reactions of wood members • For every action there is an equal and opposite reaction, creates a “state of rest” Two types of actions or stresses • Fiber Bending Stress (Fb)--a bending stress • Modulus of Elasticity (E)--stiffness of structure • considered as the deflection or amount of sag when structural members are given a load. • Deflection Allowances (Stiffness) • Floor = 1/360 • Roof = 1/240**Table Values**• Look up values for lumber type & grade • Normal Duration for fiber stress (Fb) • Typical consideration for floor loads • Modulus of Elasticity (E) • See Text Fb = fiber stress in bending E = Modulus of Elasticity (Stiffness)**Construction Lumber Considerations —Wood Type & Quality**• Wood Type Available in Area • Douglas Fir-Larch (North) • Hemlock-Fir (North) • Spruce-Pine-Fir (North) • Southern Pine • Wood Quality or Grade Value • Select Structural (Best) • No. 1/No. 2 (Normally specified) • No. 3 (Worst)**Fb**E DOUGLAS FIR-LARCH Required: Find the Fiber Stress in Bending and the Modulus of Elasticity of a 2x8 Douglas fir-Larch for grade No.1/No.2 (see text)**Hem-Fir**Fb E Required: Find the Fiber Stress in Bending and the Modulus of Elasticity of a 2x10 Hemlock-Fir for grade No.1/No.2**See Text**Problem #1: Span = 11’-8” Hem-Fir 1st Step--find E: E = 1.6 Joist Sizing & Spacing 2 1 2nd Step--Use E and find size to fit span * 3rd Step--find Fb value (2x8): Fb = 1,380 Value is under 1,380 so it works! 4th Step--determine if Fb works with E Solution = 2x8 @19.2OC (E controls failure--all fits) 4**See Text**Joist Sizing & Spacing 2 5 5 Problem #2: Span = 14’-9” Douglas Fir 1st Step--find E: E = 1.6 1 2nd Step--Use E and find sizes to fit span 3rd Step--find Fb value (2x10): Fb = 1,045 4th Step-determine if Fb works with E Value is over 1045 it 4 doesn’t work! 5th Step--using Fb find working column Solution: 2x10 @ 16 OC(Fb is tendency to failure) 4**Span Table (not in text)**• Loads • 40 LL • 10 DL • 1--Lumber type • 2--Lumber grade • 3-Spacing • 4--Span Solution: DF #2-2x10 @ 16” OC will span 14’-11”**Handout on Structural Analysis #1**• Use both charts in text • Remember that if all values in the “E” column apply and work then the modulus of Elasticity is the tendency of failure • If values are adjusted in the Fb row then the Fiber stress in bending is the tendency of failure**Beam Types**• 1--Solid timber beam • 2--Built-up dimensional lumber beam • 3--Glued Laminated beam • 4--Parallel strand lumber beam (PSL) • 5--Laminated veneer lumber beam (LVL) • 6--Truss I-Joist beam • 7--Box or Plywood beam • 8--Flitch beam (wood and steel) • 9--Steel beams**Beam Type—Built-up Dimensional Lumber Beam**• Dimensional lumber (2x6, 2x8, 2x10, 2x12) nailed and/or glued together • Vertical placement**Beam and Joist Attached with joist hangers**• Joist are attached to beams with metal joist hangers • What type of beam is shown?**Beam Type — Glued Laminated**• Dimensional lumber placed horizontally and glued together**Beam Type — Parallel Strand Lumber Beam**• See classroom example**Beam Type — Laminated Veneer Lumber Beam**• Laminated Veneer Lumber (LVL) • Made of ultrasonically graded douglas fir veneers with exterior adhesives under heat and pressure • 1 3/4” wide x (5 1/2 to 18”) depth**Beam Type — Truss I-Joist Beam**• Laminated or Solid wood (top and bottom chords) • OSB or Plywood web**Beam Type — Box or Plywood Beam**• 2x @ 12” or 16” structure with plywood skin • Designed by architect or engineer**Beam Type — Flitch Beam**• A sandwich of wood and steel**S--**I Shape W or M Shape C--Channel Shape Beam Type — Steel Beams • S shape (American Standard shape) • Often called an I-beam • W & M shapes • Wide flange design • C shape • Channel shape**Beam Type — Steel Beams**• Drawing Callouts: • Shape, Nominal height x Weight/foot • Example: W10x25**Reaction**• Reaction is the portion of the load that is transferred to the bearing points of the beam • A simple beam reaction to a load would be at the end supports. Each end would support or be required to carry half the total load**Calculating the Reactions of a Beam**Reaction formula R = wl • Total load on beam should equal reaction loads: • 25 x 900 = 22500# • R1 = 15/2 x 900# = 6750# • R2 = 10/2 x 900# = 4500# • R3 = (15/2 + 10/2) x 900 =11250# 2 W = uniform load l = length of span W = 900 #/ linear foot Span = 10’-0” Span = 15’-0” R1 R3 R2**Simple Beam Design**• Simple beam has a uniform load evenly distributed over the entire length of the beam and is supported at each end. • Uniform load = equal weight applied to each foot of beam**Simple Beam Design**• Terminology • Joist/Rafter • Beam/Girder • Post/Column • Span • Tributary area • Conditions of Design • Uniform load over length of beam • Beam supported at each end Tributaryarea of beam 15’-0” Beam span**Simple Beam Design**• Tributary area • 16’ x 15’ = 240 sq ft • Total Load on Beam • 240 x 50#/sq ft = 12,000# • Load at each supporting end • 12,000/2 = 6000# Tributaryarea of beam 15’-0” Beam span**20’-0”**BEAM 12’-0” 10’-0” Determine the size of a Solid Wood Beam using Span Table • 1)Determine the tributary area and calculate the total load (W) for the beam 10 x 12 x 63 = 7560 TLD • Select beam size from table LL = 50# DL = 13#**7560 TLD w/ span of 12’**• Solution = 4 x 14 Beam**Reading the Steel Table**• Table values of load are given in kips • 1 kip = 1000 lbs • Shape and nominal size across the top • Weight per foot is given below designation • Span is located along the left side of table**BEAM**30’-0” 18’-0” Example of Using Steel Table • Calculate load: 18 x 30 x 60 = 32400 TLD • Select Beam W18 x 40